Find all the intersection points of the following two circles: -

Let the equation of circle is
${\displaystyle {(x-h)}^2+{(y-k)}^2=r^2}$
Where, h and k are the x and y coordinates of the center of the circle.
And r is the radius of the circle.
Find the equation of circles by given information,
${\displaystyle {(x-0)}^2+{(y-0)}^2={\left(\sqrt{13}\right)}^2}$
${\displaystyle x^2+y^2=13}$
${\displaystyle x^2+y^2-13=0}$ (1)
${\displaystyle {(x-1)}^2+{(y-(-1))}^2={\left(\sqrt{17}\right)}^2}$
${\displaystyle {(x-1)}^2+{(y+1)}^2=17}$
${\displaystyle x^2+1-2x+y^2+1+2y=17}$
${\displaystyle x^2+1-2x+y^2+1+2y-17=0}$ (2)
Multiply equation (1) by -1 and add both the equation:
${\displaystyle -x^2-y^2+13=0}$
After adding:
${\displaystyle 1-2x+1+2y-4=0}$
${\displaystyle -2x+2y-2=0}$
${\displaystyle -x+y-1=0}$
${\displaystyle y=x+1}$
Substitute the value of y in equation (1),
${\displaystyle x^2+y^2-13=0}$
${\displaystyle x^2+{(x+1)}^2-13=0}$
${\displaystyle x^2+x^2+1+2x-13=0}$
${\displaystyle 2x^2+2x-12=0}$
${\displaystyle x^2+x-6=0}$
${\displaystyle x^2+3x-2x-6=0}$
${\displaystyle x^2+3x-2x-6=0}$
${\displaystyle x(x+3)-2(x+3)=0}$
${\displaystyle (x+3)(x-2)=0}$
${\displaystyle x=-3,2}$
${\displaystyle x_1=-3,x_2=2}$
${\displaystyle y_1=-3+1}$
${\displaystyle y_1=-2}$
${\displaystyle y_2=2+1}$
${\displaystyle y_2=3}$

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